Re: [PATCH v2] sched/fair: Fix insertion in rq->leaf_cfs_rq_list

[Vincent Guittot]

On Wed, 30 Jan 2019 at 15:27, Vincent Guittot
<vincent.guittot@xxxxxxxxxx> wrote:
>
> On Wed, 30 Jan 2019 at 15:01, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
> >
> > On Wed, Jan 30, 2019 at 03:01:04PM +0100, Peter Zijlstra wrote:
> > > --- a/kernel/sched/fair.c
> > > +++ b/kernel/sched/fair.c
> > > @@ -282,13 +282,15 @@ static inline struct cfs_rq *group_cfs_r
> > >       return grp->my_q;
> > >  }
> > >
> > > -static inline void list_add_leaf_cfs_rq(struct cfs_rq *cfs_rq)
> > > +static inline bool list_add_leaf_cfs_rq(struct cfs_rq *cfs_rq)
> > >  {
> > >       struct rq *rq = rq_of(cfs_rq);
> > >       int cpu = cpu_of(rq);
> > >
> > >       if (cfs_rq->on_list)
> > > -             return;
> > > +             return rq->tmp_alone_branch == &rq->leaf_cfs_rq_list;
> >
> > And I'm almost certain that can be: return true, but got my brain in a
> > twist.
>
> Yes this can return true
>
> If cfs_rq->on_list) then a child not already on the list used the path :
>
> if (cfs_rq->tg->parent &&
>            cfs_rq->tg->parent->cfs_rq[cpu]->on_list) {
>
> which does rq->tmp_alone_branch = &rq->leaf_cfs_rq_list;

In fact tests show that we must keep:
  return rq->tmp_alone_branch == &rq->leaf_cfs_rq_list;

Because the 1st sched_entity that will be used in the newly added
for_each_sched_entity loop, can be the sched_entityof the cfs_rq that
we just added in the list so cfs_rq->on_list == 1 but we must continue
to add parent

Apart from that, tests are ok

>
> >
> > > +
> > > +     cfs_rq->on_list = 1;
> > >
> > >       /*
> > >        * Ensure we either appear before our parent (if already